gaurav200x said:
1) Let the bulb be sold at x and tube light at y.
So 1.15 x = 20 and 0.85 y = 20
Net profit/loss will be (20-x)+(20-y)
he will incur a net loss of 360/391
The calculation is very tedious. If anyone has got a simple way to calculate, please let me know.
2) I dint get any of the choices, so please let me know if my approach is correct or not.
20% of the girls passing the exam + x% of the boys passing the exam = 40% of the total students of the school
i.e. 20% of 60% of total students + x% of 40% of total = 40 % of total..
i.e. 0.12 + x% of 0.4 = 0.4
i.e. x% = 0.28/0.4
=> x= 40 %
3) The eqn u get is 2/3 [2/3 {2/3 (x-1) -1} -1 ]
= 8/27 x - 38/27
Put x = 25, u get a number divisible by 3
4) answer is 64 = (2^3)^2
5) answer = -3
x^3000 > 2^4000
or x^3000 > 16^1000
keeping x = -3
27^1000> 16^1000 which is true !!
Hence the solutions...
u r rite...dude...in Q.No. 1, 3 & 4,
check out the soloutions below for all the 5 questions...
NOTE : SYMBOL '#' INDICATES MULTIPLICATION.
1) @gaurav, ur method , as u said, was tedious. Here, the simple method u can follow...
In such a case, there is always an overall loss and the absolute value of the loss.
= 2x{2}y/100{2} - x{2}
= 2 # 15{2} # 20/(100-15){2}
= 225 # 40/115 # 85
= 9 # 40/ 23 # 17
= 360/391
Hence,
1]
2) Suppose, ther r 100 students in the school; then 40 of them being boys, 60 r girls.
No. of successful students = 40.
No of successful girls = 20/100 # 60 = 12.
Therefore, boys who passed the exam (in %) = 40-12/40 # 100 = 70.
Hence,
3]
3)This method is tedious, too.
if ne1 has simple method for this, plz solve this...
Suppose 3 boys stolen x mangoes.
No. of mangoes left after the first boy's theft = 2/3 (x-1)
No. of mangoes left after the 2nd boy's theft
= 2/3 [2/3(x-1) -1]
= 4/9(x-1) - 2/3
No. of mangoes left after the 3rd boy's theft
= 2/3[4/9(x-1) - 2/3 -1]
= 8/27 (x-1) - 10/9
= 8x/27 - 38/27
This no., being divisible by 3, 8x/27 - 38/27 = 3a (for some integer a).
i.e. 8x - 38 =81a
i.e. x = 81a + 38/8
Thus, there can be an infinity of solutions, depending on the value of a.
In particular for a = 2, x= 25.
Hence,
1.
(i hav doubt on right solution for this...but this is quiet better solution)
4) Since the no. is a perfect square & a perfect cube, the no. will be in the form of x{6}.
The minimum value of x{6} (x{6} > 1) will be [when x = 2]
x{6} = 2{6} = 64.
Rite answer @ gaurav...
Also, u can solve this by trying options...Hence,
3]
5) @gaurav, slight mistake !!!
x{3000} > 2{4000} i.e. x{3} > 2{4}
i.e. x{3} > 16.
Therefore, x{3} can have a least value of 27 i.e. 3{3}
Thus, x = 3.
Hence, 2.
See the options.
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